Question

In an amusement park, a visitor gets to ride on three different rides (A, B and C) for free. On a particular day 77 opted for ride A, 55 opted for B and 50 opted for C; 25 visitors opted for both A and C, 22 opted for both A and B, while no visitor opted for both B and C. 40 visitors did not opt for ride A or B (i.e., they were outside $A \cup B$). How many visited the amusement park on that day?

• A. 102
• B. 115
• C. 130
• D. 135
• E. 150

Hint:

When a question tells you how many are {outside} \(A \cup B\), jump straight to \(N = |A \cup B| + \text{(outside)}\). Compute \(|A \cup B|\) via inclusion–exclusion and add the given outside count.

Answer 1

Answer: (E)

Step 1: Use the given “40 did not opt for A or B”.
Let the total number of visitors be \(N\). \[ N - |A \cup B| = 40 \;\Rightarrow\; N = 40 + |A \cup B|. \] Step 2: Compute \(|A \cup B|\).
\[ |A \cup B| = |A| + |B| - |A \cap B| = 77 + 55 - 22 = 110. \] Step 3: Get \(N\).
\[ N = 40 + 110 = \boxed{150}. \] (Optional check with ride C).
Given \(|B \cap C| = 0\) (so triple intersection \(=0\)): \[ |A \cup B \cup C| = 77 + 55 + 50 - 22 - 25 - 0 = 135. \] Hence \(N - |A \cup B \cup C| = 150 - 135 = 15\) visitors took none, and among the 40 outside \(A \cup B\), exactly \(25\) took only \(C\) (since \(50 - 25 - 0 = 25\)). This is consistent. \[ \boxed{\text{Total visitors } = 150} \]